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User blog:MachineGunSuper/Another Continuation to the listaments
This is a continuation to this blog post, but I thought I would make a new one just because it was getting messy there. Last time we left off at the Table of Listament Tables All Possible Tables Well, we seem to have reached a problem. How do we get bigger numbers? There is only one table of listament tables possible. Sure, we can amplify it but you may find that "lame". Well, we can create listament tables that are CLEARLY not listed on the table of listament tables. Here's how: We can apply other functions to the listament tables and to the table of listament tables other than just multiplication. Let's just start with addition. TL1 + 0.0000.......000001 TL1 + 0.0000........000002 ....... TL1 + 0.9 TL1 + 0.99 ....... TL1 + 1 TL1 + 1.000000......00001 H-Hold on a second. What are we doing? We are describing an uncountable infinity. And the main subject of this blog post: All tables of tables It has an uncountably infinite many elements, therefore it CANNOT be listed in the forms that we previously discussed. In the language of the listaments it is notated with the symbol: Ћ Ћ = {LTL+0.000000.....0000001, LTL+0.000000....0000002,................., LTL+1, LTL+1.00000.....000001. , .........., LTL+2, .......... LTL * 0.00000.....000001,......} While what I wrote above seems confusing, we can break it down into smaller pieces to help you understand it. Ћ = {LTL+0.0000....00001, LTL+ 0.000....0002, .......... , LTL+1, LTL+1.000000.....00001,......... , LTL + n} U {LTL* 0.00000...00001, LTL * 0.00000....000002,....................., LTL* n} U..... An important rule: When applying any function to a listament table, or a table of listament tables, what we are actually doing is applying that function to all it's elements. Eg: TL1^2 = ? TL1 = Listt! (1), Listt! (2), Listt! (3),..... Listt! (3), Listt! (4), Listt! (5),...... Listt! (5), Listt! (6), Listt! (7),..... TL1^2 = Listt! (1)^2, Listt! (2)^2, Listt! (3)^2,..... Listt! (3)^2, Listt! (4)^2, Listt! (5)^2,...... Listt! (5)^2, Listt! (6)^2, Listt! (7)^2,..... LTL^3 = ? TL1^3, TL2^3, TL3^3,.... TL3^3, TL4^3, TL5^3,.... TL5^3, TL6^3, TL7^3,.... Another important rule:Ћ will only contain functions that are normally used to make the original number bigger, so NO subtraction or division. Another important rule: Even though Ћ has uncountably infinite elements, it CANNOT be plugged into any listament function, because of all these 0's in the beginning, we would need an infinite number in the subscript, which won't work.That's why there is such thing as: The whole number all table of tables.It is notated with Ћ(a;b;c). Ћ(a;b;c) = {LTL+1, LTL+2, LTL+3,......., LTL *1, LTL*2, LTL*3,........, LTL^1, LTL^2, LTL^3,.........}, where the functions are listed in order from slowest growing to fastest growing. Eg: from addition to Busy Beaver and beyond. Let's apply what we've learnt... We already know how to amplify tables, so let's use other functions. Eg. addition TL1+3 = Listt! (1)+3, Listt! (2)+3, Listt! (3)+3,..... Listt! (3)+3, Listt! (4)+3, Listt! (5)+3,...... Listt! (5)+3, Listt! (6)+3, Listt! (7)+3,..... The function that this defines is : List1+n(x) In this case, it's +3 Eg: List1+3(5) = Listt! (13)+3 ~ 11 PT (54052638827.23235) TL1*5 = ? Listt! (1)*5, Listt! (2)*5, Listt! (3)*5,..... Listt! (3),*5 Listt! (4)*5, Listt! (5)*5,...... Listt! (5)*5, Listt! (6)*5, Listt! (7)*5,.... The function that this defines is List1*n(x) In this case it's *5 Eg: List1+n(3) = Listt! (7)*5 ~ 101010101.8 * 1014401 ''' '''It should really, really, ''really ''be noted that amplifying a table with a number is not, not ''not ''the same as multiplying the same table with the same number. TL1^6 = ? Listt! (1)^6, Listt! (2)^6, Listt! (3)^6,..... Listt! (3),^6 Listt! (4)^6, Listt! (5)^6,...... Listt! (5)^6, Listt! (6)^6, Listt! (7)^6,.... The function that this defines is List1^n(x) In this case it's ^6 Eg: List1^6(4) = Listt! (10)^6 = 8 PT (20175268.75698203) Surprisingly, the Ћ(a;b;c) function is not that much stronger than the ones listed above (pun intended) ListЋabcx,y,z(n). You pick the a-th element from Ћ(a;b;c) , the b-th element from the LTL, the c-th element from the TL, and the final result is repeated n times. Eg: ListЋabc5,4,2(3) = The fifth element from Ћ, which is LTL+5, the fourth element from diagonalization in LTL+5 which is TL10+5, the second element from diagonalization in TL10+5 ''' '''Just like we write the first function (the listament function) as Listt! (n), let the 10-t function to be written as Listt!({n}), which is going to be repeating 9-th function. TL10+5 = ''' '''Listt!({1})+5, Listt!({2})+5, Listt!({3})+5, Listt!({4})+5,..... Listt!({3})+5, Listt!({4})+5, Listt! ({5})+5, Listt!({6})+5,..... Listt!({5})+5, Listt!({6})+5, Listt! ({7})+5, Listt! ({8})+5,.... So ListЋabc5,4,2(3) = Listt!({4})+5 repeated 3 times = ''' '''Listt!({Listt!({Listt!({4})})}) + 5 + 5 + 5 Category:Blog posts